Question

Please help me! For algebra 2 class. 45 points!

Using the Fundamental Theorem of Algebra, complete the following exercises. Show your work.

Determine how many, what type, and find the roots for f(x) = x4 + 21x2 − 100.
Determine how many, what type, and find the roots for f(x) = x3 − 5x2 − 25x + 125.
The following graph shows a seventh-degree polynomial:
(see attachment)
Part 1: List the polynomial’s zeroes with possible multiplicities.
Part 2: Write a possible factored form of the seventh degree function.

Without plotting any points other than intercepts, draw a possible graph of the following polynomial:
f(x) = (x + 8)3 (x + 6)2 (x + 2)(x − 1)3 (x − 3)4 (x − 6).

Answer 1

1) There are four roots, with two real and two imaginary roots

The roots are x ±2, ±5·i

2) Therefore, there are 2 roots

The roots are real

The roots are x = -5 and x = 5

3) The possible factored form is therefore; (x + 5)²×(x + 1)×(x - 4)³×(x + 7)

4) Please see the attached graph of the function drawn with Microsoft Excel

Explanation:

1) The given equations is as follows;

f(x) = x⁴ + 21·x² - 100

Let a = x², we have;

f(x) = a² + 21·a - 100

a = (-21 ± √(21² - 4 × 1 ×(-100))/(2 × 1)

a = -25 or 4

Therefore, x = √4 = ±2 or x = √(-25) = ±5·i

There are four roots, with two real and two imaginary roots

2) For f(x) = x³ - 5·x² - 25·x + 125

We have;

f(x) = x³ - 5·x² - 25·x + 125

From the equation, we see that x = 5 is a solution of the equation, therefore;

f(5) = 5³ - 5·5² - 25·5 + 125 = 0

Which gives, (x - 5) is a factor of the equation,

Dividing x³ - 5·x² - 25·x + 125 by (x - 5) gives;

x² - 25

(x³ - 5·x² - 25·x + 125)/(x - 5)

x³ - 5·x²

`{}`- 25·x

`{}` -25·x + 125

`{}` 0 + 0

Therefore, by long division, (x³ - 5·x² - 25·x + 125)/(x - 5) = x² - 25

(x - 5)×(x² - 25) = (x - 5) × (x - 5) × (x + 5)

Therefore, there are 2 roots

The roots are real

The roots are x = -5 and x = 5

3) From the polynomial zeros, we have x = -5, x = -1, x = 4, and x = 7

At x = -5 the polynomial touches the x-axis given two real roots with (x + 5)² being a factor

With the root at x = -1, a factor is (x - 1)

With the root at x = 4 which has the shape of a cubic function, we have a factor of (x - 4)³

For the root at x = 1, the factor is taken as (x + 7)

The possible factored form is therefore; (x + 5)²×(x + 1)×(x - 4)³×(x + 7)

4) The given function is therefore;

f(x) = (x + 8)³(x + 6)²(x + 2)(x - 1)³(x - 3)⁴(x - 6)

(x + 8)³(x + 6)²(x + 2)(x - 1)³(x - 3)⁴(x - 6)

Please see the attached graph of the function drawn with Microsoft Excel